Question

A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium carbonate precipi­tates. The concentration of barium ion, Ba2+, in solution after reaction is_________.

Answer 1

Answer:

[Ba^2+] = 0.160 M

Explanation:

First, let's calculate the moles of each reactant with the following expression:

n = M * V

moles of K2CO3 = 0.02 x 0.200 = 0.004 moles

moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles

Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.

Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3

As you can see, 0.04 moles of  K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of

0.012 - 0.004 = 0.008 moles of Ba(NO3)2

These moles are in total volume of 50 mL (30 + 20 = 50)

So finally, the concentration of Ba in solution will be:

[Ba] = 0.008 / 0.050 = 0.160 M